$\sin(x) \leq \sum_{k=0}^n (-1)^k \frac{x^{2k+1}}{(2k+1)!}$ when $n$ is even and $x\geq 0$

Let $n\geq 2$ be an even integer and $x\geq 0$. I want to show that$$\sin(x) \leq \sum_{k=0}^n (-1)^k \frac{x^{2k+1}}{(2k+1)!}.$$

Assume first that $x\in (0,\pi]$.By Taylor's theorem with Lagrange form of the remainder, there is some $\xi\in (0,x)$ such that$$\sin(x) = \sum_{k=0}^n (-1)^k \frac{x^{2k+1}}{(2k+1)!} + \frac{\sin^{(2n+2)}(\xi)x^{(2n+2)}}{(2n+2)!}.$$

Since $n$ is even and $\xi\in [0,\pi]$, $\sin^{(2n+2)}(\xi) = -\sin(\xi) \leq 0$, and the proof is over.

How can I deal with the case where $x>\pi$ ? Perhaps the following estimate holds:$$\sum_{k=0}^n (-1)^k \frac{x^{2k+1}}{(2k+1)!}\geq 1,$$but I can't manage to show this even when $n=2$.